Answer
$\approx 1.6760$
Work Step by Step
We will solve the given integral by using u-substitution method.
Let us consider that $u=\ln (x) \implies du=\dfrac{1}{x}dx$
$\displaystyle \int_{1}^2 x \ln (2x) \ dx= \ln (2x) (\dfrac{x^2}{2}) -\int_1^2 (\dfrac{x^2}{2}) (\dfrac{1}{x}) \ dx$
or, $= [\ln (2x) (\dfrac{x^2}{2}) - (\dfrac{x^2}{4})]_1^2$
or, $=[2 \ln 4 -1]-[\dfrac{1}{2} \ln (2) -\dfrac{1}{4}]$
or, $=2 \ln 4 -\dfrac{1}{2} \ln (2)-\dfrac{3}{4}$
or, $\approx 1.6760$
